Question

Assume that 50 kg of (a) alum and (b) ferrous sulfate and lime as Ca(OH)₂ is added per 4000 m³ of wastewater. Also assume that all insoluble and very slightly soluble products of the reactions, with the exception of 15 g/m³ CaCO₃, are precipiated as sludge. How many kg of sludge /1000 m³ will result in each case?

Answer 1

Given that; 50 kg of alum and ferrous sulfate and lime as Ca(OH)₂ is added per 4000 m³

50 kg = 50000 g in 4000m^3.

15 g/m³ CaCO₃ is soluble means 15 g CaCO3 is soluble per cubic meter.

Now we calculate the soluble amount in 1000 cubic meter as follows:

15 g/m³ x   1000 m³= 15000 g soluble .

Now calculate the total insoluble and soluble sludge in 1000 cubic meter as follows:

50000 g in 4000m^3 x 1000 m^3= 125000 g the total insoluble and soluble sludge

Now calculate the insoluble sludge in 1000 cubic meter as follows:

(total insoluble and soluble sludge - the soluble amount)/1000 m^3

= 125000 g-15000 g)/1000 m^3* 1.0 kg/1000 g

= 110 kg /m^3